Temperature of one mole of helium gas is increased by 4- Thus- increase in internal energy is

Name: NEET - Grade 11 - Chemistry - Hydrocarbons - Session 09 - W06
Uploaded: 2022-07-04T10:36:42+05:30
Description: The correct option is B 12 calC_v (heat capacity at constant volume) = nbsp;(Δ EΔ T ) For monoatomic gas, C_v = 3R2 nbsp; n = 1 mole Δ E = nC_v Δ T = 32RΔ ...

The correct option is B 12 calC_v (heat capacity at constant volume) = nbsp;(Δ EΔ T ) For monoatomic gas, C_v = 3R2 nbsp; n = 1 mole Δ E = nC_v Δ T = 32RΔ ...

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Standard XII

Chemistry

Heat Capacity at Constant Volume

Temperature o...

Question

Temperature of one mole of helium gas is increased by $4^{\circ}$ . Thus, increase in internal energy is

6 cal

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12 cal

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3 cal

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24 cal

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Solution

The correct option is B 12 cal
$C_{v}$ (heat capacity at constant volume) $= (\frac{Δ E}{Δ T})$
For monoatomic gas, $C_{v} = \frac{3 R}{2}$
n = 1 mole
$Δ E = n C_{v} Δ T = \frac{3}{2} R Δ T = \frac{3}{2} \times 2 \times 4 = 12 c a l$

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Heat Capacity at Constant Volume

Standard XII Chemistry

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Temperature of one mole of helium gas is increased by 4∘. Thus, increase in internal energy is

The correct option is B 12 calCv (heat capacity at constant volume) =(ΔEΔT) For monoatomic gas, Cv=3R2 n = 1 mole ΔE=nCvΔT=32RΔT=32×2×4=12 cal

Temperature of one mole of helium gas is increased by $4^{\circ}$ . Thus, increase in internal energy is

The correct option is B 12 cal
$C_{v}$ (heat capacity at constant volume) $= (\frac{Δ E}{Δ T})$
For monoatomic gas, $C_{v} = \frac{3 R}{2}$
n = 1 mole
$Δ E = n C_{v} Δ T = \frac{3}{2} R Δ T = \frac{3}{2} \times 2 \times 4 = 12 c a l$